Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $q = \dfrac{z - 7}{-9z - 72} \div \dfrac{z^2 + 3z - 70}{z + 10} $
Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{z - 7}{-9z - 72} \times \dfrac{z + 10}{z^2 + 3z - 70} $ First factor the quadratic. $q = \dfrac{z - 7}{-9z - 72} \times \dfrac{z + 10}{(z - 7)(z + 10)} $ Then factor out any other terms. $q = \dfrac{z - 7}{-9(z + 8)} \times \dfrac{z + 10}{(z - 7)(z + 10)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (z - 7) \times (z + 10) } { -9(z + 8) \times (z - 7)(z + 10) } $ $q = \dfrac{ (z - 7)(z + 10)}{ -9(z + 8)(z - 7)(z + 10)} $ Notice that $(z + 10)$ and $(z - 7)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ \cancel{(z - 7)}(z + 10)}{ -9(z + 8)\cancel{(z - 7)}(z + 10)} $ We are dividing by $z - 7$ , so $z - 7 \neq 0$ Therefore, $z \neq 7$ $q = \dfrac{ \cancel{(z - 7)}\cancel{(z + 10)}}{ -9(z + 8)\cancel{(z - 7)}\cancel{(z + 10)}} $ We are dividing by $z + 10$ , so $z + 10 \neq 0$ Therefore, $z \neq -10$ $q = \dfrac{1}{-9(z + 8)} $ $q = \dfrac{-1}{9(z + 8)} ; \space z \neq 7 ; \space z \neq -10 $